# Discrete rational rotations on the two dimensional torus

It is well known (Kronecker's Theorem) that "irrational rotations" are dense on $[0,1)$. That is, the set $$\{ x+nr\mod 1 : n \in \mathbb{N} \}$$ is dense on $[0,1)$, provided that $r$ is irrational. This theorem is relatively easy to prove.

On the two dimensional torus $\mathbb{T}=[0,1)\times[0,1)$ (with opposite edges identified), the following result is true. The set $$\{ (x+nr \mod 1,x+nr' \mod 1) \in \mathbb{T} : n \in \mathbb{N} \}$$ is dense in $\mathbb{T}$ if and only if $\{r, r', 1\}$ are rationally independent (i.e., if there exist integers $a$ and $b$ such that $ar+br'$ is an integer, then $a=b=0$). I have seen a very complicated proof of this. Is there an "easy" proof? That is, something that one could assign for reading to an undergraduate (say a senior)?

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I took the liberty of adding a IMHO more appropriate tag. I have only seen this result in number-theoretic contexts. Actually I'm a bit curious to see how this is used in dynamical systems! – Jyrki Lahtonen Jun 6 '12 at 20:56
@Jyrki: For a completely integrable Hamiltonian system, the Arnold–Liouville theorem says that the phase space (of dim $2n$, say) is foliated into invariant tori (of dim $n$) such that the motion on each torus is just a straight line (in suitable coordinates). Whether such a trajectory is periodic or fills the torus densely depends on the rational depencence or independence of the components of the direction vector of the line. – Hans Lundmark Jun 7 '12 at 6:51
@Hans, thanks for that bit. Can't say I'm familiar with the result, but at least I can sort of see what's going on there. But here it looks like "time" (or whatever is the parameter of the motion) is ticking in discrete steps? – Jyrki Lahtonen Jun 7 '12 at 7:39
@Jyrki: True... Let's put it this way then: translation on a torus is a simple example of a discrete-time dynamical system, and it's interesting as an illustration of what type of global behaviour that orbits can exhibit. See for example Section 1.2 in Zehnder's book. – Hans Lundmark Jun 7 '12 at 8:51
@Hans, I'm afraid I get a "no eBook available". – Jyrki Lahtonen Jun 7 '12 at 9:02